\begin{problem*}{11.39}
A wad of sticky clay with mass $m$ and velocity $\vect{v}_i$ is fired
at a solid cylinder of mass $M$ and radius $R$ (Fig.~P11.39).  The
cylinder is initially at rest and is mounted on a fixed horizontal
axle that runs through its center of mass.  The line of motion of the
projectile is perpendiuclar to the axle and at a distance $d<R$ from
the center.  \Part{a} Find the angular speed of the system just after
the clay strikes and sticks to the surface of the cylinder.  \Part{b}
Is the mechanical energy of the clay--cylinder system constant in this
process?  Explain your answer.  \Part{c} Is the momentum of the
clay--cylinder system constant in this process?  Explain your answer.
\end{problem*}

\begin{solution}
\Part{a}
Conserving angular momentum
\begin{align}
  L_i = (md^2)\cdot\p({\frac{v_i}{d}})
    &= L_f = \p({mR^2+\frac{1}{2}MR^2})\cdot\omega \\
  \omega &= \frac{mdv_i}{mR^2 + \frac{1}{2}MR^2}
    = \ans{\frac{2mv_id}{(2m+M)R^2}} \;.
\end{align}

\Part{b}
Mechanical energy is not constant, because some kinetic energy is lost
to internal energy during the collision.
\begin{equation}
  \frac{K_f}{K_i}
    = \frac{\frac{1}{2}(mR^2+\frac{1}{2}MR^2)\omega^2}{\frac{1}{2}mv_i^2}
    = \frac{(2m_M)R^2\p({\frac{2mv_id}{(2m+M)R^2}})^2}{2mv_i^2}
    = \frac{2md^2}{(2m+M)R^2} \;.
\end{equation}

\Part{c}
\ans{Angular momentum is conserved} because there are no external
torques (we used this in \Part{a}).  \ans{Linear momentum is not
conserved} because the axle exerts a backwards force on the cylinder
as the clay strikes.  The axle cannot exert any net vertical force
though, because the vertical component of the clay--cylinder momentum
is zero both before and after the collision.
\end{solution}
